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Difference between revisions of "2021 AMC 12B Problems/Problem 19"

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==Problem==
 
==Problem==
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Two fair dice, each with at least <math>6</math> faces are rolled. On each face of each dice is printed a distinct integer from <math>1</math> to the number of faces on that die, inclusive. The probability of rolling a sum of <math>7</math> is <math>\frac34</math> of the probability of rolling a sum of <math>10,</math> and the probability of rolling a sum of <math>12</math> is <math>\frac{1}{12}</math>. What is the least possible number of faces on the two dice combined?
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Two fair dice, each with at least <math>6</math> faces are rolled. On each face of each die is printed a distinct integer from <math>1</math> to the number of faces on that die, inclusive. The probability of rolling a sum of <math>7</math> is <math>\frac34</math> of the probability of rolling a sum of <math>10,</math> and the probability of rolling a sum of <math>12</math> is <math>\frac{1}{12}</math>. What is the least possible number of faces on the two dice combined?
  
 
<math>\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18\qquad \textbf{(D) }19 \qquad \textbf{(E) }20</math>
 
<math>\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18\qquad \textbf{(D) }19 \qquad \textbf{(E) }20</math>
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<cmath>(a-12)(b-12)=12</cmath>
 
<cmath>(a-12)(b-12)=12</cmath>
 
This narrows the possibilities down to 3 ordered pairs of <math>(a,b)</math>, which are <math>(13,24)</math>, <math>(6,10)</math>, and <math>(8,9)</math>.  We can obviously ignore the first pair and test the next two straightforwardly.  The last pair yields the answer:
 
This narrows the possibilities down to 3 ordered pairs of <math>(a,b)</math>, which are <math>(13,24)</math>, <math>(6,10)</math>, and <math>(8,9)</math>.  We can obviously ignore the first pair and test the next two straightforwardly.  The last pair yields the answer:
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<cmath>\frac{6}{72}=\frac{3}{4}\frac{9+8-9}{72}</cmath>
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<cmath>\frac{6}{72}=\frac{3}{4}\left(\frac{9+8-9}{72}\right)</cmath>
 
The answer is then <math>a+b=8+9=\boxed{\textbf{(B)}\ 17}</math>.
 
The answer is then <math>a+b=8+9=\boxed{\textbf{(B)}\ 17}</math>.
  
 
~Hyprox1413
 
~Hyprox1413
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 +
==Solution 3 (Logic)==
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Notice that
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<cmath>
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\begin{align*}
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    7&=1+6 \\
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    &\ \ \ \ \ \ \ \ \ \vdots \\
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    &=6+1
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\end{align*}
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</cmath>
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are the only cases that could possibly form 7. In another words, regardless of the number of faces for each dice, 6 is the number of cases that could form a rolling sum of 7.
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 +
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Because the value of the total combination for both are the same, we could infer that the number of cases of having a rolling sum of 10 is <math>\frac{4}{3}\cdot6=8</math>. With the number 8, we could also deduce that one dice has 8 sides and the other has at least 9 sides. Thence trial and error could be utilized.
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 +
 +
Since 9 is the next smallest number, the case could be tested.
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<cmath>
 +
\begin{align*}
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    12&=3+9 \\
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    &=4+8 \\
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    &\ \ \ \ \ \ \ \ \ \vdots \\
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    &=8+4
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\end{align*}
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</cmath>
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<math>\frac{6}{8\cdot9}=\frac{1}{12}</math> is also true. Therefore, <math>8+9=\boxed{\textbf{(B) }17}</math>
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 +
 +
~MaPhyCom
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 +
== Video Solution ==
 +
https://youtu.be/xGp5yQ5Bshs
 +
 +
~MathProblemSolvingSkills
 +
 +
 +
  
 
== Video Solution by OmegaLearn (Using Probability) ==
 
== Video Solution by OmegaLearn (Using Probability) ==

Latest revision as of 04:26, 5 June 2025

Problem

Two fair dice, each with at least $6$ faces are rolled. On each face of each die is printed a distinct integer from $1$ to the number of faces on that die, inclusive. The probability of rolling a sum of $7$ is $\frac34$ of the probability of rolling a sum of $10,$ and the probability of rolling a sum of $12$ is $\frac{1}{12}$. What is the least possible number of faces on the two dice combined?

$\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18\qquad \textbf{(D) }19 \qquad \textbf{(E) }20$

Solution 1

Suppose the dice have $a$ and $b$ faces, and WLOG $a\geq{b}$. Since each die has at least $6$ faces, there will always be $6$ ways to sum to $7$. As a result, there must be $\tfrac{4}{3}\cdot6=8$ ways to sum to $10$. There are at most nine distinct ways to get a sum of $10$, which are possible whenever $a,b\geq{9}$. To achieve exactly eight ways, $b$ must have $8$ faces, and $a\geq9$. Let $n$ be the number of ways to obtain a sum of $12$, then $\tfrac{n}{8a}=\tfrac{1}{12}\implies n=\tfrac{2}{3}a$. Since $b=8$, $n\leq8\implies a\leq{12}$. In addition to $3\mid{a}$, we only have to test $a=9,12$, of which both work. Taking the smaller one, our answer becomes $a+b=9+8=\boxed{\textbf{(B)}\ 17}$.

Solution 2

Suppose the dice have $a$ and $b$ faces, and WLOG $a\geq{b}$. Note that if $a+b=12$ since they are both $6$, there is one way to make $12$, and incrementing $a$ or $b$ by one will add another way. This gives us the probability of making a 12 as \[\frac{a+b-11}{ab}=\frac{1}{12}\] Cross-multiplying, we get \[12a+12b-132=ab\] Simon's Favorite Factoring Trick now gives \[(a-12)(b-12)=12\] This narrows the possibilities down to 3 ordered pairs of $(a,b)$, which are $(13,24)$, $(6,10)$, and $(8,9)$. We can obviously ignore the first pair and test the next two straightforwardly. The last pair yields the answer: \[\frac{6}{72}=\frac{3}{4}\left(\frac{9+8-9}{72}\right)\] The answer is then $a+b=8+9=\boxed{\textbf{(B)}\ 17}$.

~Hyprox1413

Solution 3 (Logic)

Notice that \begin{align*} 7&=1+6 \\ &\ \ \ \ \ \ \ \ \ \vdots \\ &=6+1 \end{align*} are the only cases that could possibly form 7. In another words, regardless of the number of faces for each dice, 6 is the number of cases that could form a rolling sum of 7.


Because the value of the total combination for both are the same, we could infer that the number of cases of having a rolling sum of 10 is $\frac{4}{3}\cdot6=8$. With the number 8, we could also deduce that one dice has 8 sides and the other has at least 9 sides. Thence trial and error could be utilized.


Since 9 is the next smallest number, the case could be tested. \begin{align*} 12&=3+9 \\ &=4+8 \\ &\ \ \ \ \ \ \ \ \ \vdots \\ &=8+4 \end{align*} $\frac{6}{8\cdot9}=\frac{1}{12}$ is also true. Therefore, $8+9=\boxed{\textbf{(B) }17}$


~MaPhyCom

Video Solution

https://youtu.be/xGp5yQ5Bshs

~MathProblemSolvingSkills



Video Solution by OmegaLearn (Using Probability)

https://youtu.be/geEDrsV5Glw

~ pi_is_3.14

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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